整数的函数

已通过测试,#includevoidinput(intb[],intn);intf(inta[],intn);main(){inta[20],n,d;scanf("%d",&n);input(a,n)

输入:2212522-11119输出:Number21252ofdigit2:3Number-1111ofdigit9:0如果是上面的意思的话,代码如下:#include"stdio.h"intmai

#includemain(){inta,c,b,d;scanf("%d%d",a,b);c=a+b;d=a*b;printf("%d%d",c,d);}再问：采用函数的方法再答：先输入两个数，然后执行

#include<stdio.h>int getmax(int a,int b){      re

#includevoidprnint(inta){if(a/10==0)printf("%c",a+'0');else{prnint(a/10);printf("%c",a%10+'0');}}int

如果数据在A1,公式可为=roundup(A1,)

#includemain(){inta,b;printf("输入整数a,b:");scanf("%d,%d",&a,&b);c=func(a,b);printf("a*a+b*b=%d\n",c);}

楼主你好,这是源程序,是用辗转相除法写的#includeintmax(intp_a,intp_b){\x09if(p_b==0)\x09\x09returnp_a;\x09else\x09\x09re

voidmain(){inti,n;scanf("%d",&n);for(i=0;i*i

=IF(H1="","",IF(H1=1,TEXT(20,"0.00"),IF(H1=2,TEXT(20,"0.000"),TEXT(20,"0.0000"))))这人没有反应

intmax_common_divisor(inta,intb){//最大公约数intlarge_num,small_num,r;if(a>b){large_num=a;small_num=b;}el

#includevoidfun(intm,intn){\x05printf("%d\n",m*m+n*n);}main(){\x05intm,n;\x05while(scanf("%d%d",&m,&

a=Int(InputBox("a="))b=Int(InputBox("b="))c=Round(Rnd()*(a-b)+b,0)MsgBox"a="&a&";b="&b&";c="&c

源码如下：#includeintsum(inta,intb){\x09returna+b;\x09}intmain(){\x09intm=1,n=2,k=3;\x09intt=sum(sum(m,n)

intflip(intin)//不用数组{intout=0,flag=0;if(in>-10&&inreturnin;elseif(in{flag=1;in=-in;}do{out+=in%10;ou

#include"stdio.h"voidmain(){\x05intnum1,num2,temp,a;\x05printf("pleaseinputtwonumbers:\n");\x05scanf

//最大公倍数public:calculate(int,int);//构造函数intcmostlardivisor();//计算最大公约数intcmostlitmultiple();//计算最大公倍数

我只会MATLAB,假设数组记为x,数组中元素各不相同[a,b]=max(x);%a是最大值,b是最大值的下标,即序号x(b)=[];%删除x的最大值次大值=max(x);%次小值类似[a,b]=mi

#includeintis_prime(intn){inti=0;for(i=2;i*i

完整程序如下：#includefun(intx,inty){intr;if(x>y){x=x;y=y;}r=x;x=y;y=r;r=x%y;while(r!=0){x=y;y=r;r=x%y;}ret